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\begin{document}

\title{高等代数二}
\subtitle{8-4-对称变换和对称矩阵 }
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2023年3月9日} }

\maketitle

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.4.i. 作业：星期天晚上十点半之前在网络教学平台提交 }

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\begin{enumerate}
\item   整理课堂笔记，补充没写完的计算或证明。
\item   习题(8.4)\#1,2,3,5,6, 抄写题目。
\end{enumerate}

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\begin{frame}{8.4.ii. 目录 }

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\begin{enumerate}

\item[8.4.1.] 对称变换的定义
\item[8.4.2.] 对称变换的例子
\item[8.4.4.] 定理8.4.1. 对称变换关于规范正交基的矩阵是对称矩阵
\item[8.4.5.] 定理8.4.2. 对称变换的充分条件
\item[8.4.6.] 定理8.4.3. 实对称阵的特征值都是实数
\item[8.4.7.] 定理8.4.4. 对称变换的不同本征值的本征向量相互正交
\item[8.4.8.] 定理8.4.5. 对称变换关于某个规范正交基的矩阵是对角矩阵
\item[8.4.9.] 定理8.4.6. 实对称阵正交相似于对角矩阵
\item[8.4.10.] 例子

\end{enumerate}

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\begin{frame}{8.4.iii. 课堂讲解重点 }

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\begin{enumerate}
\item  对称变换的概念和例子
\item  对称变换关于规范正交基的矩阵
\item  实对称阵的特征值都是实数
\item  实对称阵的不同特征值的特征向量相互正交
\item  实对称阵正交相似于对角矩阵
\end{enumerate}

\end{frame}

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\begin{frame}{8.4.1. 对称变换的定义 }

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\begin{itemize}

\item  {\color{red}问题：欧氏空间 $V$ 上的线性变换 $\sigma:V\to V$ 什么时候称为对称变换？}

\item  解答：如果对任意向量 $\alpha,\beta\in V$, 等式
$$\langle\sigma(\alpha),\beta\rangle = \langle\alpha,\sigma(\beta)\rangle$$ 
都成立，那么称 $\sigma$ 是对称变换。 

\end{itemize}

\end{frame}

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\begin{frame}{8.4.2. 对称变换的例子 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：设 $V=\mathbb{R}^2$ 是行向量空间。对任意向量 $\alpha=(x_1,x_2)$ 和 $\beta=(y_1,y_2)$, 其内积定义为 $\langle\alpha,\beta\rangle = x_1y_1+x_2y_2=\alpha\cdot\beta^t$. 
设矩阵\, {\footnotesize $A=\begin{pmatrix}2&3\\ 3&5 \end{pmatrix}$}, 设线性变换 $\sigma:V\to V$ 由 $\sigma(\alpha)=\alpha\cdot A$ 所定义。验证这是对称变换。
}

\item  解答：
\begin{eqnarray*}
\langle\sigma(\alpha),\beta\rangle &=& \langle\alpha\cdot A,\beta\rangle = \alpha\cdot A\cdot \beta^t, \\ 
\langle\alpha,\sigma(\beta)\rangle &=& \langle\alpha,\beta\cdot A\rangle = \alpha\cdot (\beta\cdot A)^t = \alpha\cdot A^t \cdot \beta^t. 
\end{eqnarray*}
%分别取 $\alpha,\beta$ 为标准基的向量 $(1,0),(0,1)$, 可得 $A=A^t$. 
因为 $A=A^t$, 所以有
\begin{eqnarray*}
\langle\sigma(\alpha),\beta\rangle = \langle\alpha,\sigma(\beta)\rangle. 
\end{eqnarray*}
因此 $\sigma$ 是对称变换。

\end{itemize}

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%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{8.4.3. 对称变换的例子}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：设 $V$ 是二维的欧氏空间。设线性变换 $\sigma:V\to V$ 关于某个正交基下的矩阵是对角阵。证明这个线性变换是对称变换。}

\item  证明：
\begin{enumerate}
\item  设线性变换 $\sigma$ 关于正交基 $\Phi=\{\alpha_1,\alpha_2\}$ 的矩阵是对角阵\,  
{\footnotesize 
$A=\begin{pmatrix} \lambda_1 &0 \\  0&\lambda_2 \\  \end{pmatrix}$.
}
\item  由线性变换的矩阵的定义得 
$\sigma(\alpha_1)=\lambda_1\alpha_1, 
\sigma(\alpha_2)=\lambda_2\alpha_2$. 

\item  因为 $\Phi$ 是正交基，可得对任意向量 $\alpha,\beta\in \Phi$, 都有 
$\langle\sigma(\alpha),\beta\rangle = \langle\alpha,\sigma(\beta)\rangle$. 

\item  因为 $\Phi$ 是正交基，可得对任意向量 $\alpha,\beta\in V$, 都有
$\langle\sigma(\alpha),\beta\rangle = \langle\alpha,\sigma(\beta)\rangle$. 

 \end{enumerate}
 
\end{itemize}

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\begin{frame}{8.4.4. 定理8.4.1. 对称变换在规范正交基下的矩阵}

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%每页详细内容

\begin{itemize}

\item  {\color{red}定理：设 $\sigma:V\to V$ 是 $n$ 维欧氏空间 $V$ 上的对称变换。则 $\sigma$ 在任意规范正交基下的矩阵是对称矩阵。}

\item  证明：
\begin{enumerate}
\item  设 $\Phi=\{\alpha_1, \alpha_2,\cdots,\alpha_n\}$ 是 $V$ 的规范正交基。 
\item  由规范正交基的定义可得 $\langle \Phi^t,\Phi \rangle = E_n$. 
\item  根据线性变换关于一个基的矩阵的定义，设 $\sigma(\Phi)=\Phi\cdot A$. 
\item  一方面， $\langle \sigma(\Phi)^t,\Phi \rangle = \langle (\Phi\cdot A)^t,\Phi \rangle 
= \langle A^t\cdot\Phi^t,\Phi \rangle = A^t \langle \Phi^t,\Phi \rangle = A^tE_n=A^t$. 
\item  另一方面，$\langle \Phi^t, \sigma(\Phi) \rangle = \langle \Phi^t,\Phi\cdot A \rangle 
= \langle \Phi^t,\Phi \rangle A = E_nA=A$. 
\item  因为 $\sigma$ 是对称变换，所以 $\langle \sigma(\Phi)^t,\Phi \rangle = \langle \Phi^t, \sigma(\Phi) \rangle$. 
\item  因此 $A^t=A$. 
\end{enumerate}

\end{itemize}

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\begin{frame}{8.4.5. 定理8.4.2. 对称变换的充分条件}

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%每页详细内容

\begin{itemize}

\item  {\color{red}定理：设 $\sigma:V\to V$ 是欧氏空间 $V$ 上的线性变换。如果 $\sigma$ 在某个规范正交基下的矩阵是对称矩阵，那么 $\sigma$ 是对称变换。}

\item  证明：
\begin{enumerate}
\item  设 $\Phi=\{\alpha_1, \alpha_2,\cdots,\alpha_n\}$ 是符合定理前提的某个规范正交基。 
\item  根据线性变换关于一个基的矩阵的定义，设 $\sigma(\Phi)=\Phi\cdot A$. 
\item  根据题目条件，矩阵 $A$ 是对称矩阵，即 $A^t=A$. 
\item  由规范正交基的定义可得 $\langle \Phi^t,\Phi \rangle = E_n$. 
\item  一方面， $\langle \sigma(\Phi)^t,\Phi \rangle = \langle (\Phi\cdot A)^t,\Phi \rangle 
= \langle A^t\cdot\Phi^t,\Phi \rangle = A^t \langle \Phi^t,\Phi \rangle = A^tE_n=A^t$. 
\item  另一方面，$\langle \Phi^t, \sigma(\Phi) \rangle = \langle \Phi^t,\Phi\cdot A \rangle 
= \langle \Phi^t,\Phi \rangle A = E_nA=A$. 
\item  因为 $A^t=A$, 所以 $\langle \sigma(\Phi)^t,\Phi \rangle = \langle \Phi^t, \sigma(\Phi) \rangle$. 
\item  因此 $\sigma$ 是对称变换。
\end{enumerate}

\end{itemize}

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\begin{frame}{8.4.6. 定理8.4.3. 实对称阵的特征值都是实数 }

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\begin{itemize}

\item  {\color{red}问题：}
\begin{enumerate}
\item  {\color{red}证明二阶实对称阵的特征值都是实数。}
\item  {\color{red}证明 $n$ 阶实对称阵的特征值都是实数。}
\end{enumerate}

\item  证明：
\begin{enumerate}
\item  直接计算特征多项式 
$$f(x) = \det(xE-A)=\begin{vmatrix} x-a&-b \\ -b&x-c \\  \end{vmatrix} = x^2-(a+c)x+ac-b^2. $$
这个二次多项式的判别式为 $\Delta = (a-c)^2+4b^2\ge 0$. 故有两个实根。 %所以矩阵 $A$ 有两个实数特征值。

\item  设在复数范围内有 $AX=\lambda X$. 转置共轭得 $\bar{X}^tA=\bar{\lambda} \bar{X}^t$. 考察下述计算，
$$\bar{\lambda} (\bar{X}^tX) = (\bar{\lambda} \bar{X}^t)X = (\bar{X}^tA)X = \bar{X}^t(AX) = \bar{X}^t(\lambda X) = \lambda  (\bar{X}^tX). $$

\end{enumerate}

\end{itemize}

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\begin{frame}{8.4.7. 定理8.4.4. 对称变换的本征向量}

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\begin{itemize}

\item  {\color{red}定理：对称变换的属于不同本征值的本征向量彼此正交。}

\item  证明：
\begin{enumerate}
\item  设 $\sigma$ 是对称变换。
\item  设有 $\sigma(\alpha_1)=\lambda_1\alpha_1$ 与 $\sigma(\alpha_2)=\lambda_2\alpha_2$. 
\item  因为 $\sigma$ 是对称变换，所以 
$ \langle\sigma(\alpha_1),\alpha_2 \rangle = \langle\alpha_1,\sigma(\alpha_2)\rangle $. 
\item  两边计算可得 $\langle \lambda_1\alpha_1,\alpha_2 \rangle = \langle\alpha_1,\lambda_2\alpha_2 \rangle $. 
\item  由内积运算法则可得 $\lambda_1 \langle\alpha_1,\alpha_2 \rangle = \lambda_2 \langle\alpha_1,\alpha_2 \rangle $. 
\item  因为 $\lambda_1\neq \lambda_2$, 所以 $\langle\alpha_1,\alpha_2 \rangle = 0$. 
\item  因此 $\alpha_1$ 与 $\alpha_2$ 相互正交。
\end{enumerate}

\end{itemize}

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\begin{frame}{8.4.8. 定理8.4.5. 对称变换的矩阵}

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%每页详细内容

\begin{itemize}

\item   {\color{red}定理：设 $V$ 是一个 $n$ 维欧氏空间。设 $\sigma:V\to V$ 是一个对称变换。证明存在一个规范正交基，使得这个对称变换的矩阵是对角矩阵。}

\item  证明：对维数 $n$ 使用归纳法。
\begin{enumerate}
\item  因为 $\sigma$ 是对称变换，所以存在实本征值 $\lambda_1$ 与单位长度的实本征向量 $\alpha_1$.  
\item  考虑一维的向量子空间 $W=L(\alpha_1)$, 以及 $n-1$ 维的正交补空间 $W^{\,\perp}$. 
\item  设任意 $\xi\in W^{\,\perp}$ 与任意 $\eta\in W$, 因为 $\sigma$ 是对称变换，所以有 

\vspace{-0.4cm}
$$\langle\sigma(\xi),\eta \rangle = \langle \xi,\sigma(\eta) \rangle = \langle \xi,\lambda_1\eta \rangle= \lambda_1\langle \xi,\eta \rangle=0. $$
\vspace{-0.6cm}

所以 $\sigma(\xi)\in W^{\,\perp}$. 因此 $W^{\,\perp}$ 是 $\sigma$ 的不变子空间。
\item  限制的线性变换 $\sigma\mid_{W^{\,\perp}}$ 仍是对称变换，所以根据归纳假设，存在 $W^{\,\perp}$ 的规范正交基 $\{\alpha_2,\cdots,\alpha_n\}$, 使得这个限制变换关于这个基的矩阵是对角阵

\vspace{-0.4cm}
$$
\text{diag}\{\lambda_2,\cdots,\lambda_n\}. 
 $$
\vspace{-0.6cm}

\item  因此 线性变换 $\sigma$ 关于规范正交基 $\{\alpha_1,\alpha_2,\cdots,\alpha_n\}$ 的矩阵是对角阵 

\vspace{-0.4cm}
$$
\text{diag}\{\lambda_1, \lambda_2,\cdots,\lambda_n\}. 
 $$
\vspace{-0.6cm}

\end{enumerate}


\end{itemize}

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\begin{frame}{8.4.9. 定理8.4.6. 实对称阵正交相似于对角矩阵 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}定理：设 $A$ 是实对称矩阵。则存在正交矩阵 $U$ 使得 $U^{\,t}AU$ 是对角阵。}

\item  证明：
\begin{enumerate}
\item  计算矩阵 $A$ 的特征多项式。
\item  实对称矩阵的所有特征值都是实数。
\item  对每个特征值，计算线性无关的特征向量。
\item  对同一个特征值的特征向量，使用正交化方法，得到正交的特征向量。
\item  不同特征值的特征向量已经相互正交。
\item  将相互正交的特征向量规范化，按列向量的方式排列成一个矩阵 $U$. 
\item  因为 $U$ 的列向量组是规范正交基，所以 $U$ 是正交矩阵，即有 $U^t=U^{-1}$. 
\item  所以 $U^{\,t}AU = U^{-1}AU$ 是对角线元素为相应的特征值的对角阵。

\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：设 $A$ 是下述矩阵，求正交矩阵 $U$ 使得 $U^{\,t}AU$ 是对角阵，
{\footnotesize 
$$A=\begin{pmatrix} 4&2&2 \\ 2&4&2 \\ 2&2&4  \end{pmatrix}.$$
}
}

\item  解答：
\begin{enumerate}
\item  计算特征多项式 $f(x)=\det(xE-A)=(x-2)^2(x-8)$. 
\item  计算属于 $x=2$ 的特征向量 $\eta_1,\eta_2$, 进行正交化和规范化，得到 $\gamma_1,\gamma_2$. 
\item  计算属于 $x=8$ 的特征向量 $\eta_3$, 进行规范化，得到 $\gamma_3$. 
\item  所求正交基为 $U=(\gamma_1,\gamma_2,\gamma_3)$, 得到对角阵为 $\text{diag}\{2,2,8\}$. 
\end{enumerate}


\end{itemize}

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\begin{frame}{习题(8.4)\#1 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：设 $\sigma$ 是 $n$ 维欧氏空间 $V$ 上的一个线性变换，证明如果 $\sigma$ 满足下述三个条件中的任意两个条件，那么 $\sigma$ 必然满足第三个条件：
\begin{enumerate}[(a)]
\item  $\sigma$ 是正交变换；
\item  $\sigma$ 是对称变换；
\item  $\sigma$ 是对合变换，即 $\sigma^2=\iota$ 是恒等变换。
\end{enumerate}

}

\item  思路：设线性变换 $\sigma$ 关于某个规范正交基的矩阵是 $A$, 则上述条件成为
\begin{enumerate}[(a)]
\item  $A^t=A^{-1}$. 
\item  $A^t=A$. 
\item  $A^{-1}=A$. 
\end{enumerate}

\end{itemize}

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\begin{frame}{习题(8.4)\#2 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：
设 $\sigma$ 是 $n$ 维欧氏空间 $V$ 上的一个对称变换，且 $\sigma^2=\sigma$. 
证明存在 $V$ 的一个规范正交基，使得 $\sigma$ 关于这个基的矩阵为 
{\footnotesize 
\begin{eqnarray*}
A=\begin{pmatrix} E_r & O \\ O & O \end{pmatrix}. 
\end{eqnarray*}
}
}

\item  思路：应用实对称矩阵正交相似于对角矩阵这一定理。

\end{itemize}

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\begin{frame}{习题(8.4)\#3 }

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\begin{itemize}

\item  {\color{red}问题：证明两个对称变换的和仍是一个对称变换。
}

\item  思路：根据对称变换的定义。或者证明关于矩阵的相应结论。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：如果矩阵 $A$ 是满足条件 $A^t=-A$, 那么称 $A$ 是反对称矩阵。
证明实反对称矩阵的特征值要么是零，要么是纯虚数。
}

\item  思路：仿照实对称阵的类似结论的证明。

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：对下述实数矩阵 $A$, 求正交矩阵 $U$, 使得 $U^tAU$ 是对角矩阵，
{\footnotesize 
\begin{eqnarray*}
(i)\,\,\, A=\begin{pmatrix} 11&2&-8 \\ 2&2&10 \\ -8&10&5 \end{pmatrix}; 
\hspace{1cm}
(ii)\,\,\, A=\begin{pmatrix} 17&-8&4 \\ -8&17&-4 \\ 4&-4&11 \end{pmatrix}. 
\end{eqnarray*}
}
}

\item  思路：先计算特征值和特征向量。然后从特征向量求出一个规范正交基。


\end{itemize}

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